KCET · Physics · Wave Optics
A coil of inductive reactance \( \frac{1}{\sqrt{3}} \Omega \) and resistance \( 1 \Omega \) is connected to a \( 200 \mathrm{~V}, 50 \mathrm{~Hz} \) ac.
supply. The time lag between maximum voltage and current is
- A \( \frac{1}{300} s \)
- B \( \frac{1}{600} s \)
- C \( \frac{1}{500} \mathrm{~s} \)
- D \( \frac{1}{200} s \)
Answer & Solution
Correct Answer
(B) \( \frac{1}{600} s \)
Step-by-step Solution
Detailed explanation
Given, inductive reactance, \( \omega L=\frac{1}{\sqrt{3}} \Omega \); resistance, \( R=1 \Omega \); ac voltage, \( V=200 \mathrm{~V}, 50 \mathrm{~Hz} \);
Now, \( \tan \phi=\frac{\omega L}{R} \)
\[
\begin{array}{l}
\Rightarrow \tan \phi=\frac{1}{\sqrt{3}} \\
\Rightarrow \phi=30^{\circ} \text { or } \phi=\frac{\Pi}{6} \\
\Rightarrow \omega t=\frac{\Pi}{6} \\
\Rightarrow t=\frac{\Pi}{6} \times \frac{1}{\omega}
\end{array}
\]
Since \( \omega=2 \Pi f=2 \Pi \times 50 \)
\( t=\frac{\Pi}{6} \times \frac{1}{2 \Pi \times 50}=\frac{1}{600} s \)
Therefore, time lag between maximum voltage and current is \( 1 / 600 \mathrm{~s} \).
Now, \( \tan \phi=\frac{\omega L}{R} \)
\[
\begin{array}{l}
\Rightarrow \tan \phi=\frac{1}{\sqrt{3}} \\
\Rightarrow \phi=30^{\circ} \text { or } \phi=\frac{\Pi}{6} \\
\Rightarrow \omega t=\frac{\Pi}{6} \\
\Rightarrow t=\frac{\Pi}{6} \times \frac{1}{\omega}
\end{array}
\]
Since \( \omega=2 \Pi f=2 \Pi \times 50 \)
\( t=\frac{\Pi}{6} \times \frac{1}{2 \Pi \times 50}=\frac{1}{600} s \)
Therefore, time lag between maximum voltage and current is \( 1 / 600 \mathrm{~s} \).
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